Whilst thinking about this question, I came across a problem. The original question was for what differentiable functions $f:\mathbb{R} \to \mathbb{R}$, with $y=f(x)$, does $\frac{dy}{dx} = F(y)$ for some function $F$; note the domain of $F$ is the image of $f$. Injectivity is sufficient, letting $F = f' \circ f^{-1}$. If $f$ is constant, then, even though $f^{-1}$ is not well defined, we can still define the derivative as a function of $y$ in a natural manner, namely $F(y) = 0$. How to characterize this leads me to the question.

Consider $f:\mathbb{R} \to \mathbb{R}$ differentiable with the following property. For the fiber $S_a$ of some $a$ in the image of $f$, $f'(x) = c$ for constant $c$ for all $x \in S_a$. If this holds for every $a$ in the image, then either $f$ is injective or $f$ is constant.

Roughly speaking, $f$ has the property that $a=f(x_1)=f(x_2)=\cdots f(x_n)$ implies $c=f'(x_1)=f'(x_2)=\cdots f'(x_n)$ (roughly because the set which maps to $a$ needn't be countable).

Perhaps this is trivial. I can't seem to find a counterexample.

  • $\begingroup$ I don't understand the question. What role does $f$ play? Do we have $y = f(x)$? $\endgroup$ – Qiaochu Yuan Jul 12 '16 at 3:36
  • $\begingroup$ @QiaochuYuan Concerning the principal portion in the orange section, $f$ is the function, $y$ is a fixed element in the image of $f$. (though perhaps the confusion stems from the fact that, in the first paragraph, I did have $y=f(x)$). $\endgroup$ – MathematicsStudent1122 Jul 12 '16 at 3:37

Consider $$ f(x)=\begin{cases} x^3&\text{for }x<0,\\ 0&\text{for }0\leq x\leq 1,\\ (x-1)^3&\text{for }x>1. \end{cases} $$ For $a\in\mathbb R$ nonzero, $f^{-1}(a)$ is a singleton so your condition holds. On the other hand $f^{-1}(0)=[0,1]$ and $f'(x)=0$ on $[0,1]$. It's possible to find examples that are infinitely differentiable if desired.

However $f(x)$ is quite constrained by the conditions. Firstly if $|f^{-1}(y)|>1$ then $f'=0$ on $f^{-1}(y)$. Suppose otherwise, and pick $b>a$ with $f(a)=f(b)=y$. wlog suppose $f'(a)>0$. Then $f(x)>y$ for $x\in(a,a_1)$ for some $a_1>a$. Let $$ c=\inf\{x\in(a,b]\mid f(x)=y\}. $$ Then $c\in[a_1,b]$ and $f(c)=y$ by continuity. Thus $f'(c)=f'(a)>0$, so $f(x)<y$ for $x\in(c_1,c)$ for some $c_1<c$. Pick $a_2\in(a,a_1)$, $c_2\in(c_1,c)$ with $a_2<c_2$. Then $f(a_2)>y$, $f(c_2)<y$ and $f(x)\neq y$ for $x\in[a_2,c_2]$. This contradicts the IVT.

Secondly $f$ is monotonic (either nonincreasing or nondecreasing). Suppose otherwise, and wlog suppose $a<b<c$ with $f(a)\leq f(c)<f(b)$. By the MVT, $f'(d)<0$ for some $d\in(b,c)$. However the IVT implies $f(d_1)=f(d)$ for some $d_1\in(a,b)$. This contradicts the first property.

Therefore $f$ is either nondecreasing or nonincreasing, and if there are any $y$-values whose preimage is a nontrivial interval (a "plateau"), then $f'$ is zero at the ends of the plateau.

  • $\begingroup$ This is technically correct, in that it contradicts the statement, but I was hoping for something more "exotic" than a mere piecewise constant example $\endgroup$ – MathematicsStudent1122 Jul 13 '16 at 3:11
  • $\begingroup$ @MathematicsStudent1122 The IVT implies that any counterexample will have $f'(x)=0$ whenever $|f^{-1}(f(x))|>1$. Hence $f$ is either nondecreasing or nonincreasing. So they all look something like this. $\endgroup$ – stewbasic Jul 13 '16 at 3:32
  • $\begingroup$ @MathematicsStudent1122 I added proofs for the claims in my comment. $\endgroup$ – stewbasic Jul 13 '16 at 4:20
  • $\begingroup$ Updated answer is perfect. Thank you. $\endgroup$ – MathematicsStudent1122 Jul 15 '16 at 2:56

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